Design Article

IMG1

Gain error affects op amp choices

Soufiane Bendaoud, National Semiconductor

7/14/2006 12:04 PM EDT

Some applications, such as test automation, signal conditioning in process control, and instrumentation, require the amplification of small signals. Open-loop gain can be a determining factor in the choice of the operational amplifier (op amp) for an application such as a data acquisition system. However, it is important to consider the error term associated with gain. To better understand and visualize this nonideality of the amplifier, we will write out the transfer functions and manipulate the mathematical expressions to isolate the error term or gain error.

As an example, let's consider the configuration of Figure 1:



Ideally,

Vout/Vin = 1 + R2/R1. This is true if we consider the op amp to be ideal. But there's more to that than just this simple expression when considering nonideal op amps. The gain is expressed in the form of Equation 1:



Vout/Vin = A(s)/(1+A(s)β.

β, in this case, is R1/(R1+R2), and where A(s) is the open-loop gain and β is the feedback factor, also defined as the portion of the output being fed back to the input.

Dividing the numerator and denominator of Equation 1 by A(s) gives the following expression: 1/(1/A(s) + β).

We now multiply this expression by 1/β and obtain Equation 2:



To simplify Equation 2 above, we write T = A(s) β.

We then add and subtract the same term T to the numerator of Equation 2.

Rearranging the equation yields Equation 3:



It becomes apparent from Equation 3 that the first term 1/b is the ideal gain and that the second term, 1 − 1/(1 + A(s)β) is the fractional gain error.

When A(s) approaches infinity, the gain is simply equal to the ideal gain 1/β. When A(s) is less than infinity, however, we must consider the error term. Table 1 depicts an example of various op amps configured in a gain of 100 (β = 1/100) under the same conditions, but with different open-loop gains.





Clearly, the error term is proportional to noise gain (closed-loop gain), but inversely proportional to open-loop gain. Thus it's important to choose a high open-loop gain amp for applications requiring high closed-loop gains.

About the author
Soufiane Bendaoud is a product marketing manager at National Semiconductor Corp. National Semiconductor Corp., Sunnyvale, CA.


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Comments


manu-s

10/20/2009 2:54 PM EDT

The error in the first case when AB is 178 should be much less. In the equation 3 we have to take modulus of the denominator - so
ACL = 1/b*(1/(1/sqrt[(AB)^2+1] - because if we don't do that then when AB=1 we get ACL = 1/2B which is not correct because that's the 3db point and at that point we get ACL = 1/(sqrt(2)*B).So when AB=178 then error = 0.01% .

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